THEORY EXAMINATION (SEM–VI) 2016-17 THEORY OF MACHINES-II

THEORY OF MACHINES – II (EME603)

Section-wise Solved Answers & Notes

SECTION – A (10 × 2 = 20 Marks)

Very short, direct answers

(a) Damping factor

Given:   Natural frequency fn=1f_n = 1fn​=1 Hz
Damped frequency fd=0.9f_d = 0.9fd​=0.9 Hz

fd=fn1−ζ2f_d = f_n \sqrt{1-\zeta^2}fd​=fn​1−ζ2​ 0.9=11−ζ2⇒ζ=1−0.81=0.4350.9 = 1 \sqrt{1-\zeta^2} \Rightarrow \zeta = \sqrt{1-0.81} = \boxed{0.435}0.9=11−ζ2​⇒ζ=1−0.81​=0.435​ 

(b) Definitions

• Stability: Ability of governor to return to equilibrium after disturbance
• Sensitiveness: Ratio of mean equilibrium speed to speed range
• Isochronism: All radii corresponding to same speed
• Hunting: Continuous oscillation of governor balls

(c) Friction circle

Friction circle is a circle drawn at pin joint whose radius equals frictional resistance divided by normal reaction, used in mechanism force analysis.

(d) Need of large flywheels in punching/shearing

Because these operations require high energy for short duration, flywheel supplies stored energy and smoothens speed fluctuations.

(e) Absorption vs Transmission dynamometer

(f) Open & Cross belt drive

• Open belt: Pulleys rotate in same direction
• Cross belt: Pulleys rotate in opposite directions

(g) Gyroscopic couple vs Reaction couple

(h) Gyroscopic couple on ship

Given:
I = 2000 kg-m²
N = 360 rpm
v = 30 km/h = 8.33 m/s
R = 200 m

ω=2πN60=37.7 rad/s\omega = \frac{2\pi N}{60} = 37.7 \text{ rad/s}ω=602πN​=37.7 rad/s Ω=vR=0.0417 rad/s\Omega = \frac{v}{R} = 0.0417 \text{ rad/s}Ω=Rv​=0.0417 rad/s C=IωΩ=2000×37.7×0.0417C = I \omega \Omega = 2000 \times 37.7 \times 0.0417C=IωΩ=2000×37.7×0.0417 C=3144 N-mC = \boxed{3144 \text{ N-m}}C=3144 N-m​

Effect: Ship dips at bow and rises at stern.

(i) Governor effort & power

• Governor effort: Mean force exerted at sleeve
• Governor power: Work done by sleeve movement

(j) Controlling force for Porter governor

Fc=(W+w)w×mω2rF_c = \frac{(W+w)}{w} \times \frac{m\omega^2 r}{}Fc​=w(W+w)​×mω2r​

(Diagram compulsory in exam)

SECTION – B (Any 5 × 10 = 50 Marks)

(a) Vibration of motor mounted on springs

Given:                                                          Mass = 120 kg
Armature mass = 35 kg                              Eccentricity = 0.5 mm
Speed = 1500 rpm                                      Key relations:

ω=2πN60\omega = \frac{2\pi N}{60}ω=602πN​ Transmissibility=111\text{Transmissibility} = \frac{1}{11}Transmissibility=111​ k=mωn2k = m\omega_n^2k=mωn2​

Steps:                                                          Find forcing frequency

Determine natural frequency                      Calculate stiffness of one spring

Find transmitted force                                (Show all steps clearly in exam)

(b) Proell governor equilibrium speed

Given:                                                         Ball mass = 6 kg
Central load = 150 kg                                Arm length = 200 mm
Radius = 180 mm

N=602π(W+w)gwrN = \frac{60}{2\pi}\sqrt{\frac{(W+w)g}{wr}}N=2π60​wr(W+w)g​​

Substitute values to get equilibrium speed.

(c) Gyroscopic effect on two-wheeler

Gyroscopic couple + centrifugal force causes leaning inward during turning.

Limiting speed:

v=mgRIωv = \sqrt{\frac{mgR}{I\omega}}v=IωmgR​​ 

(d) Locomotive balancing

Given data used to calculate:                  (a) Swaying couple
(b) Tractive effort variation                      (c) Hammer blow

Key formula:

Hammer blow=mrω2\text{Hammer blow} = m r \omega^2Hammer blow=mrω2 

(e) Turning moment & flywheel fluctuation

Given:

T=5000+1500sin⁡3θT = 5000 + 1500\sin3\thetaT=5000+1500sin3θ

Mean torque = 5000 N-m

Power:                                             P=2πNT60P = \frac{2\pi NT}{60}P=602πNT​

Fluctuation of speed:

Nmax−NminN×100\frac{N_{max}-N_{min}}{N} \times 100NNmax​−Nmin​​×100 

(f) Braking of vehicle on inclined plane

Given:                                             v = 36 km/h = 10 m/s
µ = 0.5

Distance:                                        s=v22as = \frac{v^2}{2a}s=2av2​

Time:

t=vat = \frac{v}{a}t=av​                  Solve separately for:
• Vehicle moving up                      • Vehicle moving down

(g) Porter governor expressions

Effort:

E=(Fmax−Fmin)×yE = (F_{max}-F_{min}) \times yE=(Fmax​−Fmin​)×y

Power:

P=E×yP = E \times yP=E×y

Coefficient of insensitiveness:

Ci=Nmax−NminNmeanC_i = \frac{N_{max}-N_{min}}{N_{mean}}Ci​=Nmean​Nmax​−Nmin​​ 

 SECTION – C (Any 2 × 15 = 30 Marks)

Q3. Hartnell governor (Spring-loaded)

Given data used to find:                          • Initial spring compression
• Speed range including friction

Key equations:                                        Fc=mω2rF_c = m\omega^2 rFc​=mω2r Fs=kx+F0F_s = kx + F_0Fs​=kx+F0​

Include:
• Obliquity correction                               • Friction force ±30 N

Q4. Steam engine dynamics

Given:
Bore = 250 mm                                      Stroke = 400 mm
Speed = 120 rpm

Steps:                                                        Calculate piston force

Turning moment on crank                        Thrust on bearings

Flywheel acceleration                                T=F×rT = F \times rT=F×r 

Q5. Complete balancing of rotating masses

Given angular positions and radii.

Steps:                                                         Draw force polygon

Draw couple polygon                                Determine mass A

Find angular position

Condition:

∑mr=0,∑mrl=0\sum mr = 0,\quad \sum mr l = 0∑mr=0,∑mrl=0